How should we pool effect sizes in meta-analysis?
2024-07-30
To get our meta-analysis result, we need to take a weighted average of effect sizes…
But how should we weigh the evidence of each study?
Study weight \(w_{k}\) is inversely related to the variance (\(SE^2_{k}\))
\[ w_{k}=\frac{1}{(SE)^{2}_{k}} \]
The pooled estimate \(\widehat{\theta}\) is the weighted average:
\[ \widehat{\theta} = \frac{\Sigma\ \widehat{\theta}_{k} w_{k}}{\Sigma\ w_{k}} \]
To compute the standard error on our averaged result \(\widehat\theta\)
\[ SE(\widehat{\theta}) = \frac{1}{\Sigma\ w} \]
. . .
Too simplistic to account for differences between studies.
The true effect size of study \(k\) (\(\theta_{k}\)) is drawn from a distribution with mean \(\mu\) with error \(\zeta_{k}\).
\[ {\theta}_{k}=\mu+\zeta_{k} \]
The reported result (\(\widehat{\theta}_{k}\)) of a study is generated by the true effect size (\(\theta_{k}\)) plus sampling error (\(\epsilon_{k}\)).
\[ \widehat{\theta}_{k}=\theta_{k} +\epsilon_{k} \]
\[ w^{*}_{k}=\frac{1}{SE^{2}_{k} + \tau^2} \]
Then compute the weighted average in the usual fashion.
Tip
My slides are online here:
https://jmoggridge.github.io/fixed-and-random-effects